Abstract
Let \(M(z)=A_mz^m+A_{m-1}z^{m-1}+\cdots +A_1z+A_0\) be a matrix polynomial, whose coefficients \(A_k\in {{\mathbb {C}}}^{n\times n}\), \(\forall \, k=0,1,\ldots , m\), satisfying the following dominant property
then it is known that all eigenvalues \(\lambda\) of M(z) locate in the open disk
In this paper, among other things, we prove some refinements of this result, which in particular provide refinements of some results concerning the distribution of zeros of polynomials in the complex plane.
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1 Introduction
Let \({\mathbb {C}}^{n \times n}\) be the set of all \(n \times n\) matrices whose entries are in \({\mathbb {C}}\). By matrix polynomial, we mean the matrix-valued function of a complex variable of the form
where \(A_i\in {\mathbb {C}}^{n\times n}\) for all \(i=0,1,2,\ldots ,m\). If \(A_m\ne 0\), M(z) is called a matrix polynomial of degree m. A number \(\lambda\) is called an eigenvalue of the matrix polynomial M(z), if there exits a nonzero vector \(X\in {\mathbb {C}}^n\), such that \(M(\lambda )X=0\). The vector X is called an Eigenvector of M(z) associated to the eigenvalue \(\lambda\). It should be noted that each finite eigenvalue of M(z) is a root of the characteristic polynomial \(\det (M(z))\). The polynomial eigenvalue problem is to find an eigenvalue \(\lambda\) and a non-zero vector \(X\in {\mathbb {C}}^n\) such that \(M(\lambda )X=0\). For \(m=1\), it is actually the generalized eigenvalue problem
and in addition, if \(B=I\), we have the standard eigenvalue problem
Computing eigenvalues of a matrix polynomial is a hard problem. There are iterative methods to compute these eigenvalues (for reference see [5]). Moreover, when computing pseudospectra of matrix polynomials, which provide information about the global sensitivity of the eigenvalues, a particular region of the (possibly extended) complex plane must be identified that contains the eigenvalues of interest, and bounds clearly help to determine such region (for details see [6]). Therefore, it is useful to find the location of these eigenvalues. Note that, if \(A_0\) is singular, then 0 is an eigenvalue of M(z), and if \(A_m\) is singular, then 0 is an eigenvalue of the matrix polynomial \(z^m M(1/z)\). Therefore, to locate the eigenvalues of these matrix polynomials, we always assume that \(A_0\) and \(A_m\) are non-singular.
Notations:
For a matrix \(A\in {\mathbb {C}}^{n\times n}\), the notation \(A\ge 0\) means "A is positive semidefnite", that is for every vector \(X \in {\mathbb {C}}^n\) we have \(X^*AX\ge 0\). By \(A>0\), we mean "A is positive definite", that is \(X^*AX> 0\) for every \(X \in {\mathbb {C}}^n\). Also in this paper for any two matrices \(A,B \in {\mathbb {C}}^{n \times n}\), the notation \(A\ge B\) means \(A-B\ge 0\). Throughout this paper,\(\left\| . \right\|\) denotes a subordinate matrix norm.
In the theory of distribution of zeros of polynomials with complex coefficients, we have the following result due to Cauchy [4, p. 123]
Theorem A
Let \(P(z)=a_0+a_1z+a_2z^2+\cdots +a_mz^m\) be a polynomial of degree m and
then all the zeros of P(z) lie in the circle \(|z|<1+{\mathcal {M}}\).
As an application of Theorem A, Dehmer [3] proved the following (also see [4, Theorem 27.2])
Theorem B
Let \(P(z)=a_0+a_1z+a_2z^2+\cdots +a_mz^m\), \(a_m\ne 0\), \(m\ge 1\) be a complex polynomial of degree m, such that \(|a_m|>|a_i|\) for all \(i=0,1,\ldots ,m-1\), then all the zeros of P(z) lie in the disk \(|z|< 2\).
Trính et al. [2] extended this result to the matrix polynomials and proved the following:
Theorem C
Let \(M(z)=A_0+A_1z+\cdots +A_mz^m\) be a matrix polynomial, whose coefficients \(A_i\in {\mathbb {C}}^{n\times n}\) satisfying the following dominant property
Then each eigenvalue \(\lambda\) of M(z) locate in the open disk
In the same paper, they proved the direct extension of Theorem A to matrix polynomial in the form of the following result:
Theorem D
Let \(M(z)=A_0+A_1z+\cdots +A_mz^m\) be a matrix polynomial of degree m, whose coefficients \(A_k\in {\mathbb {C}}^{n\times n}\).
Then each eigenvalue \(\lambda\) of M(z) satisfies
where
In this paper, we obtain an annulus containing all the eigenvalues of the matrix polynomial M(z) and use it to obtain a refinement of Theorem D. In this direction, we have the following:
Theorem 1.1
Let \(M(z)=A_mz^m+A_{m-1}z^{m-1}+\cdots +A_1z+A_0\) be a matrix polynomial of degree m, where \(A_k\in {\mathbb {C}}^{n\times n}\) and \(A_0, A_m\) are invertible. If \(\alpha _1, \alpha _2, \ldots , \alpha _m\) are m non-zero real or complex numbers, such that \(\sum _{k=1}^{m}|\alpha _k|\le 1\), then each eigenvalue \(\lambda\) of M(z) satisfies \(r_1 \le |\lambda | \le r_2\), where
and
Remark 1
If we take\(n=1\) and let\(A_i=[a_i], i=0,1,\ldots ,m\), we get a result due to Aziz and Qayoom [1] for the zeros of a polynomial with real or complex coefficients.
Now as a refinement of Theorem D, we have the following:
Theorem 1.2
Let
be a matrix polynomial of degree m, whose coefficients \(A_k\in {\mathbb {C}}^{n\times n}\) \((k=0,1,2,\ldots ,p,m)\) and let
then every eigenvalue \(\lambda\) of M(z) satisfies
For \(p=m-1\), we have the following:
Corollary 1
Let
be a matrix polynomial of degree m, whose coefficients \(A_k\in {\mathbb {C}}^{n\times n}\) \((k=0,1,2,\ldots ,m)\) and let
then every eigenvalue \(\lambda\) of M(z) satisfies
This result is a refinement of Theorem D.
Remark 2
A result of Aziz and Qayoom [1] for the distribution of zeros of polynomials with real or complex coefficients is a special case of Theorem1.2, when we take\(n=1\) and\(A_i=[a_i]\), \({forall}, i=0,1,2,\ldots , m\).
In particular, if \(\Vert A_m\Vert > \Vert A_i\Vert\), \(\forall \, i=0,1,2,\ldots ,p\), then from Theorem 1.2, we get the following:
Corollary 2
Let
be a matrix polynomial of degree m, whose coefficients \(A_k\in {\mathbb {C}}^{n\times n}\) \((k=0,1,2,\ldots ,p,m)\) and let \(\Vert A_m\Vert > \Vert A_i\Vert\), \(\forall \, i=0,1,2,\ldots ,p\), then each eigenvalue \(\lambda\) of M(z) satisfies
If in Corollary 2, we choose \(p=m-1\), we get the following:
Corollary 3
Let
be a matrix polynomial of degree m, whose coefficients \(A_k\in {\mathbb {C}}^{n\times n}\) \((k=0,1,2,\ldots ,m)\) and let \(\Vert A_m\Vert > \Vert A_i\Vert\), \(\forall \, i=0,1,2,\ldots ,m-1\), then each eigenvalue \(\lambda\) of M(z) satisfies
This result is a refinement of Theorem C.
2 Lemmas
Before proving the main results, let us first prove the following lemma:
Lemma 1
Let \(M(z)=A_mz^m+A_{p}z^{p}+\cdots +A_1z+A_0\), \(0\le p\le m-1\) be a matrix polynomial of degree m, where \(A_k\in {\mathbb {C}}^{n\times n}\) and \(A_m\) is invertible. If \(\alpha _1, \alpha _2, \ldots , \alpha _{p+1}\) are \(p+1\) non-zero real or complex numbers, such that \(\sum _{k=1}^{p+1}|\alpha _k|\le 1\), then each eigenvalue \(\lambda\) of M(z) lie in the disk
Proof
Let
This implies
or
therefore, we have
This implies
Let \(\lambda\) be any eigenvalue of M(z) and X be corresponding unit eigenvector. If possible, suppose that \(|\lambda |> r\), then we have by using (8)
Therefore \(\Vert M(\lambda )X\Vert >0\) for \(|\lambda |>r\). Hence our supposition is wrong and thus each eigenvalue \(\lambda\) of M(z) lies in
\(\square\)
3 Proofs of theorems
Proof of 1.1
We have \(M(z)=A_0+A_1z+\cdots +A_mz^m\). Let \(\lambda\) be an eigenvalue of M(z) and \(X\in {\mathbb {C}}^n\) be the corresponding eigenvector of M(z). We first show that each eigenvalue \(\lambda\) of M(z) lie in \(|\lambda |\le r_2.\)
We have
Applying the case when \(p=m-1\), we have from Lemma 1 that each eigenvalue \(\lambda\) of M(z) lies in
Hence we conclude that \(|\lambda |\le r_2\).
Now, consider the matrix polynomial
Proceeding similarly as above, we have each eigenvalue \(\lambda\) of \(M^*(z)\) satisfies
Replacing z by 1/z and noting that \(M(z)=z^mM^*\left( \frac{1}{z}\right)\), we conclude that if \(\lambda\) is an eigenvalue of M(z), then
This completes the proof of theorem.\(\square\)
Proof of 1.2
We have
therefore
Now let us choose
This implies
Since \(A_{p-k+1}=0\) for \(k=p+2,p+3,\ldots ,m\), therefore \(\alpha _k=0,\,\,\text {for}\,\, k=p+2,p+3,\ldots ,m.\) Hence from (11), we have by using (10)
That is
Now by Lemma 1 and with the help of (12), we have that each eigenvalue \(\lambda\) of M(z) satisfies
This completes the proof of theorem.\(\square\)
References
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Le, Cong-Trinh., Du. Thi-Hao-Binh, and Tran-Duc. Nguyen. 2019. On the location of eigenvalues of matrix polynomials. Operators and Matrices 13 (4): 937–954.
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Acknowledgements
We are highly thankful to the referee for his/her invaluable suggestions.
Funding
The first author acknowledges the financial support given by the Science and Engineering Research Board, Govt of India under Mathematical Research Impact-Centric Sport(MATRICS) Scheme vide SERB Sanction order No: F : MTR/2017/000508, Dated 28-05-2018.
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Shah, W.M., Singh, S. On the bounds of eigenvalues of matrix polynomials. J Anal 31, 821–829 (2023). https://doi.org/10.1007/s41478-022-00481-3
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DOI: https://doi.org/10.1007/s41478-022-00481-3